A student drives a moped along a straight road as described by the velocity vers

Question

A student drives a moped along a straight road as described by the velocity versus time graph in the figure. The divisions along the horizontal axis represent 5.0 s and the divisions along the vertical axis represent 2.0 m/s. Sketch this graph in the middle of a sheet of graph paper. (Do this on paper. You will need it to do part (a) and (b).)

(a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (Do this on paper. Your instructor may ask you to turn in your work.)

(b) Sketch a graph of the acceleration versus time directly below the velocity-time graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (Do this on paper. Your instructor may ask you to turn in your work.)

(c) What is the acceleration at t = 30.0 s?
m/s2

(d) Find the position (relative to the starting point) at t = 30.0 s.
m

(e) What is the moped's final position at t = 45.0 s?
m

Explanation / Answer

Ans:-

c) The acceleration is the slope of the line in the velocity/time graph. At t = 30.0 s, (six divisions from the origin) the slope is negative and is 4 vertical divisions divided by 2 horizontal divisions: 8/10 = 0.8 m/s²

d) The position is v*dt; this is the total area under the graph between t = 0 and t = 30. This is area is made up of 3 sections

from t = 0 to t = 15 the area is the area of a triangle of base 15 and altitude 8; the area is 0.5*15*8 =60
from t = 15 to t = 25 the area is the area of the rectangle of base 10 and height 8 = 80
from t = 25 to t = 30, the area is the area of a triangle of base 5 and altitude 4 plus the rectangle of base 5 and height 4; the area is 0.5*5*4 + 5*4 = 30

add them up to get 60 + 80 + 30 = 170 m

e) the first two parts of b) apply here. now there are two additional triangles to consider: one has a base of 10 and an altitude of 8 and the other a base of 10 and an altitude of -8. The areas are equal and opposite so cancel out, leaving the areas from the first two 40 m

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